today’s outline - november 16, 2016csrri.iit.edu/~segre/phys405/16f/lecture_22.pdf · today’s...
Post on 21-Jun-2020
5 Views
Preview:
TRANSCRIPT
Today’s Outline - November 16, 2016
• Fermi energy
• Band theory
• Dirac comb
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 21, 2016
Homework Assignment #11:Chapter 5:6,9,12,13,32,33due Monday, November 28, 2016
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 1 / 14
Today’s Outline - November 16, 2016
• Fermi energy
• Band theory
• Dirac comb
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 21, 2016
Homework Assignment #11:Chapter 5:6,9,12,13,32,33due Monday, November 28, 2016
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 1 / 14
Today’s Outline - November 16, 2016
• Fermi energy
• Band theory
• Dirac comb
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 21, 2016
Homework Assignment #11:Chapter 5:6,9,12,13,32,33due Monday, November 28, 2016
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 1 / 14
Today’s Outline - November 16, 2016
• Fermi energy
• Band theory
• Dirac comb
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 21, 2016
Homework Assignment #11:Chapter 5:6,9,12,13,32,33due Monday, November 28, 2016
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 1 / 14
Today’s Outline - November 16, 2016
• Fermi energy
• Band theory
• Dirac comb
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 21, 2016
Homework Assignment #11:Chapter 5:6,9,12,13,32,33due Monday, November 28, 2016
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 1 / 14
Today’s Outline - November 16, 2016
• Fermi energy
• Band theory
• Dirac comb
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 21, 2016
Homework Assignment #11:Chapter 5:6,9,12,13,32,33due Monday, November 28, 2016
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 1 / 14
Reciprocal space
The energy of the 3-D states is
with
kx =nxπ
lx, ky =
nyπ
ly, kz =
nzπ
lz
Enxnynz =~2π2
2m
(n2xl2x
+n2yl2y
+n2zl2z
)
=~2k2x2m
+~2k2y2m
+~2k2z2m
these discrete values of kx ,ky , kz form a lattice calledthe reciprocal lattice.
each point is a discrete singleparticle state
kx
ky
kz
1lx
1ly
1lz
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 2 / 14
Reciprocal space
The energy of the 3-D states is
with
kx =nxπ
lx, ky =
nyπ
ly, kz =
nzπ
lz
Enxnynz =~2π2
2m
(n2xl2x
+n2yl2y
+n2zl2z
)
=~2k2x2m
+~2k2y2m
+~2k2z2m
these discrete values of kx ,ky , kz form a lattice calledthe reciprocal lattice.
each point is a discrete singleparticle state
kx
ky
kz
1lx
1ly
1lz
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 2 / 14
Reciprocal space
The energy of the 3-D states is
with
kx =nxπ
lx, ky =
nyπ
ly, kz =
nzπ
lz
Enxnynz =~2π2
2m
(n2xl2x
+n2yl2y
+n2zl2z
)
=~2k2x2m
+~2k2y2m
+~2k2z2m
these discrete values of kx ,ky , kz form a lattice calledthe reciprocal lattice.
each point is a discrete singleparticle state
kx
ky
kz
1lx
1ly
1lz
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 2 / 14
Reciprocal space
The energy of the 3-D states is
with
kx =nxπ
lx, ky =
nyπ
ly, kz =
nzπ
lz
Enxnynz =~2π2
2m
(n2xl2x
+n2yl2y
+n2zl2z
)
=~2k2x2m
+~2k2y2m
+~2k2z2m
these discrete values of kx ,ky , kz form a lattice calledthe reciprocal lattice.
each point is a discrete singleparticle state
kx
ky
kz
1lx
1ly
1lz
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 2 / 14
Reciprocal space
The energy of the 3-D states is
with
kx =nxπ
lx,
ky =nyπ
ly, kz =
nzπ
lz
Enxnynz =~2π2
2m
(n2xl2x
+n2yl2y
+n2zl2z
)
=~2k2x2m
+~2k2y2m
+~2k2z2m
these discrete values of kx ,ky , kz form a lattice calledthe reciprocal lattice.
each point is a discrete singleparticle state
kx
ky
kz
1lx
1ly
1lz
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 2 / 14
Reciprocal space
The energy of the 3-D states is
with
kx =nxπ
lx, ky =
nyπ
ly,
kz =nzπ
lz
Enxnynz =~2π2
2m
(n2xl2x
+n2yl2y
+n2zl2z
)
=~2k2x2m
+~2k2y2m
+~2k2z2m
these discrete values of kx ,ky , kz form a lattice calledthe reciprocal lattice.
each point is a discrete singleparticle state
kx
ky
kz
1lx
1ly
1lz
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 2 / 14
Reciprocal space
The energy of the 3-D states is
with
kx =nxπ
lx, ky =
nyπ
ly, kz =
nzπ
lz
Enxnynz =~2π2
2m
(n2xl2x
+n2yl2y
+n2zl2z
)
=~2k2x2m
+~2k2y2m
+~2k2z2m
these discrete values of kx ,ky , kz form a lattice calledthe reciprocal lattice.
each point is a discrete singleparticle state
kx
ky
kz
1lx
1ly
1lz
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 2 / 14
Reciprocal space
The energy of the 3-D states is
with
kx =nxπ
lx, ky =
nyπ
ly, kz =
nzπ
lz
Enxnynz =~2π2
2m
(n2xl2x
+n2yl2y
+n2zl2z
)
=~2k2x2m
+~2k2y2m
+~2k2z2m
these discrete values of kx ,ky , kz form a lattice calledthe reciprocal lattice.
each point is a discrete singleparticle state
kx
ky
kz
1lx
1ly
1lz
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 2 / 14
Reciprocal space
The energy of the 3-D states is
with
kx =nxπ
lx, ky =
nyπ
ly, kz =
nzπ
lz
Enxnynz =~2π2
2m
(n2xl2x
+n2yl2y
+n2zl2z
)
=~2k2x2m
+~2k2y2m
+~2k2z2m
these discrete values of kx ,ky , kz form a lattice calledthe reciprocal lattice.
each point is a discrete singleparticle state
kx
ky
kz
1lx
1ly
1lz
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 2 / 14
Reciprocal space
The energy of the 3-D states is
with
kx =nxπ
lx, ky =
nyπ
ly, kz =
nzπ
lz
Enxnynz =~2π2
2m
(n2xl2x
+n2yl2y
+n2zl2z
)
=~2k2x2m
+~2k2y2m
+~2k2z2m
these discrete values of kx ,ky , kz form a lattice calledthe reciprocal lattice.
each point is a discrete singleparticle state
kx
ky
kz
1lx
1ly
1lz
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 2 / 14
Fermi level
Since electrons are fermions, only two (spin up and spin down) can occupyeach single particle state.
As we fill states starting with the lowestenergies, the occupied states begin to fill an octant of a sphere in k-space.
The radius of this sphere is calledkF and the volume of the octant isgiven by
defining ρ = Nq/V as the densityof free electrons per unit volume,we have
kF defines the Fermi surface andthe corresponding Fermi energy isgiven by
1
8
(4
3πk3F
)=
Nq
2
(π3
V
)kF =
(3ρπ2
)1/3EF =
~2
2m
(3ρπ2
)2/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 3 / 14
Fermi level
Since electrons are fermions, only two (spin up and spin down) can occupyeach single particle state. As we fill states starting with the lowestenergies, the occupied states begin to fill an octant of a sphere in k-space.
The radius of this sphere is calledkF and the volume of the octant isgiven by
defining ρ = Nq/V as the densityof free electrons per unit volume,we have
kF defines the Fermi surface andthe corresponding Fermi energy isgiven by
1
8
(4
3πk3F
)=
Nq
2
(π3
V
)kF =
(3ρπ2
)1/3EF =
~2
2m
(3ρπ2
)2/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 3 / 14
Fermi level
Since electrons are fermions, only two (spin up and spin down) can occupyeach single particle state. As we fill states starting with the lowestenergies, the occupied states begin to fill an octant of a sphere in k-space.
The radius of this sphere is calledkF and the volume of the octant isgiven by
defining ρ = Nq/V as the densityof free electrons per unit volume,we have
kF defines the Fermi surface andthe corresponding Fermi energy isgiven by
1
8
(4
3πk3F
)=
Nq
2
(π3
V
)kF =
(3ρπ2
)1/3EF =
~2
2m
(3ρπ2
)2/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 3 / 14
Fermi level
Since electrons are fermions, only two (spin up and spin down) can occupyeach single particle state. As we fill states starting with the lowestenergies, the occupied states begin to fill an octant of a sphere in k-space.
The radius of this sphere is calledkF and the volume of the octant isgiven by
defining ρ = Nq/V as the densityof free electrons per unit volume,we have
kF defines the Fermi surface andthe corresponding Fermi energy isgiven by
1
8
(4
3πk3F
)=
Nq
2
(π3
V
)
kF =(3ρπ2
)1/3EF =
~2
2m
(3ρπ2
)2/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 3 / 14
Fermi level
Since electrons are fermions, only two (spin up and spin down) can occupyeach single particle state. As we fill states starting with the lowestenergies, the occupied states begin to fill an octant of a sphere in k-space.
The radius of this sphere is calledkF and the volume of the octant isgiven by
defining ρ = Nq/V as the densityof free electrons per unit volume,we have
kF defines the Fermi surface andthe corresponding Fermi energy isgiven by
1
8
(4
3πk3F
)=
Nq
2
(π3
V
)
kF =(3ρπ2
)1/3EF =
~2
2m
(3ρπ2
)2/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 3 / 14
Fermi level
Since electrons are fermions, only two (spin up and spin down) can occupyeach single particle state. As we fill states starting with the lowestenergies, the occupied states begin to fill an octant of a sphere in k-space.
The radius of this sphere is calledkF and the volume of the octant isgiven by
defining ρ = Nq/V as the densityof free electrons per unit volume,we have
kF defines the Fermi surface andthe corresponding Fermi energy isgiven by
1
8
(4
3πk3F
)=
Nq
2
(π3
V
)kF =
(3ρπ2
)1/3
EF =~2
2m
(3ρπ2
)2/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 3 / 14
Fermi level
Since electrons are fermions, only two (spin up and spin down) can occupyeach single particle state. As we fill states starting with the lowestenergies, the occupied states begin to fill an octant of a sphere in k-space.
The radius of this sphere is calledkF and the volume of the octant isgiven by
defining ρ = Nq/V as the densityof free electrons per unit volume,we have
kF defines the Fermi surface andthe corresponding Fermi energy isgiven by
1
8
(4
3πk3F
)=
Nq
2
(π3
V
)kF =
(3ρπ2
)1/3
EF =~2
2m
(3ρπ2
)2/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 3 / 14
Fermi level
Since electrons are fermions, only two (spin up and spin down) can occupyeach single particle state. As we fill states starting with the lowestenergies, the occupied states begin to fill an octant of a sphere in k-space.
The radius of this sphere is calledkF and the volume of the octant isgiven by
defining ρ = Nq/V as the densityof free electrons per unit volume,we have
kF defines the Fermi surface andthe corresponding Fermi energy isgiven by
1
8
(4
3πk3F
)=
Nq
2
(π3
V
)kF =
(3ρπ2
)1/3EF =
~2
2m
(3ρπ2
)2/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 3 / 14
Energy of a Fermi shell
We may now calculate the total en-ergy of the electron gas by consider-ing a shell of this octant with thick-ness dk
the volume of the shell is
18(4πk2)dk = 1
2πk2dk
the number of states in the shell isthen
since each state has an energy~2k2/2m, the total energy of thestates in the shell is
k
k
k z
y
x
k
dk
2
[12πk2dk]
(π3/V )=
V
π2k2dk
dE =~2k4
2m
V
π2dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 4 / 14
Energy of a Fermi shell
We may now calculate the total en-ergy of the electron gas by consider-ing a shell of this octant with thick-ness dk
the volume of the shell is
18(4πk2)dk = 1
2πk2dk
the number of states in the shell isthen
since each state has an energy~2k2/2m, the total energy of thestates in the shell is
k
k
k z
y
x
k
dk
2
[12πk2dk]
(π3/V )=
V
π2k2dk
dE =~2k4
2m
V
π2dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 4 / 14
Energy of a Fermi shell
We may now calculate the total en-ergy of the electron gas by consider-ing a shell of this octant with thick-ness dk
the volume of the shell is
18(4πk2)dk = 1
2πk2dk
the number of states in the shell isthen
since each state has an energy~2k2/2m, the total energy of thestates in the shell is
k
k
k z
y
x
k
dk
2
[12πk2dk]
(π3/V )=
V
π2k2dk
dE =~2k4
2m
V
π2dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 4 / 14
Energy of a Fermi shell
We may now calculate the total en-ergy of the electron gas by consider-ing a shell of this octant with thick-ness dk
the volume of the shell is
18(4πk2)dk = 1
2πk2dk
the number of states in the shell isthen
since each state has an energy~2k2/2m, the total energy of thestates in the shell is
k
k
k z
y
x
k
dk
2
[12πk2dk]
(π3/V )=
V
π2k2dk
dE =~2k4
2m
V
π2dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 4 / 14
Energy of a Fermi shell
We may now calculate the total en-ergy of the electron gas by consider-ing a shell of this octant with thick-ness dk
the volume of the shell is
18(4πk2)dk = 1
2πk2dk
the number of states in the shell isthen
since each state has an energy~2k2/2m, the total energy of thestates in the shell is
k
k
k z
y
x
k
dk
2
[12πk2dk]
(π3/V )=
V
π2k2dk
dE =~2k4
2m
V
π2dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 4 / 14
Energy of a Fermi shell
We may now calculate the total en-ergy of the electron gas by consider-ing a shell of this octant with thick-ness dk
the volume of the shell is
18(4πk2)dk = 1
2πk2dk
the number of states in the shell isthen
since each state has an energy~2k2/2m, the total energy of thestates in the shell is
k
k
k z
y
x
k
dk
2
[12πk2dk]
(π3/V )
=V
π2k2dk
dE =~2k4
2m
V
π2dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 4 / 14
Energy of a Fermi shell
We may now calculate the total en-ergy of the electron gas by consider-ing a shell of this octant with thick-ness dk
the volume of the shell is
18(4πk2)dk = 1
2πk2dk
the number of states in the shell isthen
since each state has an energy~2k2/2m, the total energy of thestates in the shell is
k
k
k z
y
x
k
dk
2[12πk2dk]
(π3/V )
=V
π2k2dk
dE =~2k4
2m
V
π2dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 4 / 14
Energy of a Fermi shell
We may now calculate the total en-ergy of the electron gas by consider-ing a shell of this octant with thick-ness dk
the volume of the shell is
18(4πk2)dk = 1
2πk2dk
the number of states in the shell isthen
since each state has an energy~2k2/2m, the total energy of thestates in the shell is
k
k
k z
y
x
k
dk
2[12πk2dk]
(π3/V )=
V
π2k2dk
dE =~2k4
2m
V
π2dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 4 / 14
Energy of a Fermi shell
We may now calculate the total en-ergy of the electron gas by consider-ing a shell of this octant with thick-ness dk
the volume of the shell is
18(4πk2)dk = 1
2πk2dk
the number of states in the shell isthen
since each state has an energy~2k2/2m, the total energy of thestates in the shell is
k
k
k z
y
x
k
dk
2[12πk2dk]
(π3/V )=
V
π2k2dk
dE =~2k4
2m
V
π2dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 4 / 14
Energy of a Fermi shell
We may now calculate the total en-ergy of the electron gas by consider-ing a shell of this octant with thick-ness dk
the volume of the shell is
18(4πk2)dk = 1
2πk2dk
the number of states in the shell isthen
since each state has an energy~2k2/2m, the total energy of thestates in the shell is
k
k
k z
y
x
k
dk
2[12πk2dk]
(π3/V )=
V
π2k2dk
dE =~2k4
2m
V
π2dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 4 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk
=~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V
= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk
=~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V
= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk =
~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V
= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk =
~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V
= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk =
~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V
= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk =
~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V
= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk =
~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V
= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk =
~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V= dW
= PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk =
~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk =
~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V
=2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk =
~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Total energy of the gas
The total energy of the gas is therefore given by integrating dE out to theFermi wavenumber
Etot =~2V
2π2m
∫ kF
0k4dk =
~2k5FV10π2m
=~2(3π2Nq)5/3
10π2mV−2/3
If we now look at how the total energy depends on a small change of thevolume of the solid
dEtot
dV= −2
3
~2(3π2Nq)5/3
10π2mV−5/3
dEtot = −2
3Etot
dV
V= dW = PdV
this is equivalent to a pressure ex-erted by the gas which does workand changes the energy of the gas
P = −2
3
Etot
V=
2
3
~2k5F10π2m
=(3π2)2/3~2
5mρ5/3
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 5 / 14
Band theory
A better model is one which includes the periodic potential of the positiveions in a solid.
This can be demonstrated by asimple 1-D model called the Diraccomb but is generalizable to anypotential with the same mathemat-ical characteristics.
The Dirac comb is a periodic po-tential, that is
where a is the distance between therepeating potential features
we can define an operator D, calledthe displacement operator, whichdescribes the translational symme-try of the potential
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
V (x + a) = V (x)
H = − ~2
2m
d2
dx2+ V (x)
D f (x) = f (x + a)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 6 / 14
Band theory
A better model is one which includes the periodic potential of the positiveions in a solid.
This can be demonstrated by asimple 1-D model called the Diraccomb but is generalizable to anypotential with the same mathemat-ical characteristics.
The Dirac comb is a periodic po-tential, that is
where a is the distance between therepeating potential features
we can define an operator D, calledthe displacement operator, whichdescribes the translational symme-try of the potential
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
V (x + a) = V (x)
H = − ~2
2m
d2
dx2+ V (x)
D f (x) = f (x + a)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 6 / 14
Band theory
A better model is one which includes the periodic potential of the positiveions in a solid.
This can be demonstrated by asimple 1-D model called the Diraccomb but is generalizable to anypotential with the same mathemat-ical characteristics.
The Dirac comb is a periodic po-tential, that is
where a is the distance between therepeating potential features
we can define an operator D, calledthe displacement operator, whichdescribes the translational symme-try of the potential
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
V (x + a) = V (x)
H = − ~2
2m
d2
dx2+ V (x)
D f (x) = f (x + a)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 6 / 14
Band theory
A better model is one which includes the periodic potential of the positiveions in a solid.
This can be demonstrated by asimple 1-D model called the Diraccomb but is generalizable to anypotential with the same mathemat-ical characteristics.
The Dirac comb is a periodic po-tential, that is
where a is the distance between therepeating potential features
we can define an operator D, calledthe displacement operator, whichdescribes the translational symme-try of the potential
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
V (x + a) = V (x)
H = − ~2
2m
d2
dx2+ V (x)
D f (x) = f (x + a)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 6 / 14
Band theory
A better model is one which includes the periodic potential of the positiveions in a solid.
This can be demonstrated by asimple 1-D model called the Diraccomb but is generalizable to anypotential with the same mathemat-ical characteristics.
The Dirac comb is a periodic po-tential, that is
where a is the distance between therepeating potential features
we can define an operator D, calledthe displacement operator, whichdescribes the translational symme-try of the potential
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
V (x + a) = V (x)
H = − ~2
2m
d2
dx2+ V (x)
D f (x) = f (x + a)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 6 / 14
Band theory
A better model is one which includes the periodic potential of the positiveions in a solid.
This can be demonstrated by asimple 1-D model called the Diraccomb but is generalizable to anypotential with the same mathemat-ical characteristics.
The Dirac comb is a periodic po-tential, that is
where a is the distance between therepeating potential features
we can define an operator D, calledthe displacement operator, whichdescribes the translational symme-try of the potential
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
V (x + a) = V (x)
H = − ~2
2m
d2
dx2+ V (x)
D f (x) = f (x + a)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 6 / 14
Band theory
A better model is one which includes the periodic potential of the positiveions in a solid.
This can be demonstrated by asimple 1-D model called the Diraccomb but is generalizable to anypotential with the same mathemat-ical characteristics.
The Dirac comb is a periodic po-tential, that is
where a is the distance between therepeating potential features
we can define an operator D, calledthe displacement operator, whichdescribes the translational symme-try of the potential
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
V (x + a) = V (x)
H = − ~2
2m
d2
dx2+ V (x)
D f (x) = f (x + a)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 6 / 14
Band theory
A better model is one which includes the periodic potential of the positiveions in a solid.
This can be demonstrated by asimple 1-D model called the Diraccomb but is generalizable to anypotential with the same mathemat-ical characteristics.
The Dirac comb is a periodic po-tential, that is
where a is the distance between therepeating potential features
we can define an operator D, calledthe displacement operator, whichdescribes the translational symme-try of the potential
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
V (x + a) = V (x)
H = − ~2
2m
d2
dx2+ V (x)
D f (x) = f (x + a)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 6 / 14
Band theory
A better model is one which includes the periodic potential of the positiveions in a solid.
This can be demonstrated by asimple 1-D model called the Diraccomb but is generalizable to anypotential with the same mathemat-ical characteristics.
The Dirac comb is a periodic po-tential, that is
where a is the distance between therepeating potential features
we can define an operator D, calledthe displacement operator, whichdescribes the translational symme-try of the potential
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
V (x + a) = V (x)
H = − ~2
2m
d2
dx2+ V (x)
D f (x) = f (x + a)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 6 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) =
D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)
−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)
−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) =
D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)
−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)
−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) =
D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)
−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)
−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) =
D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)
−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)
−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) = D
(− ~2
2m
d2
dx2+ V (x)
)f (x)
−(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)
−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)
−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) = D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)
−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)
−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) = D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)
−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)
−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) = D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)
−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) = D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)
−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) = D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)−(− ~2
2mf ′′(x + a) + V (x)f (x + a)
)
= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) = D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)
= 0
but since V (x + a) = V (x)
, H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Properties of the displacement operator
The solution to the Dirac comb must be an eigenfunction of theHamiltonian but how will it transform under the displacement operator?
To determine this, let’s compute the commutator of D and H,[D,H] = DH − HD
[D,H] f (x) = D
(− ~2
2m
d2
dx2+ V (x)
)f (x)−
(− ~2
2m
d2
dx2+ V (x)
)Df (x)
= D
(− ~2
2mf ′′(x) + V (x)f (x)
)−(− ~2
2m
d2
dx2+ V (x)
)f (x + a)
=
(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)−(− ~2
2mf ′′(x + a) + V (x + a)f (x + a)
)= 0
but since V (x + a) = V (x), H and D commute and we can havesimultaneous eigenfunctions of both operators
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 7 / 14
Bloch functions
The eigenvalue equation for the displace-ment operator is thus
since λ cannot be zero, it must be a com-plex number which can expressed as aphasor
we thus have a condition on the wave-function in a periodic potential
resulting in Bloch’s theorem
Dψ(x) = λψ(x)
ψ(x + a) = λψ(x)
λ = e iKa
ψ(x + a)= e iKaψ(x)
Bloch’s theorem only applies to an infintely repeating potential and ourgoal is to model a macroscopic crystal so the number of repeating units ofthe potential is of the order of 1010
by assuming that the edges of the solid are always far away, and applyingperiodic boundary conditions, Bloch’s theorem can be made applicable tothe Dirac comb and other models of solids
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 8 / 14
Bloch functions
The eigenvalue equation for the displace-ment operator is thus
since λ cannot be zero, it must be a com-plex number which can expressed as aphasor
we thus have a condition on the wave-function in a periodic potential
resulting in Bloch’s theorem
Dψ(x) = λψ(x)
ψ(x + a) = λψ(x)
λ = e iKa
ψ(x + a)= e iKaψ(x)
Bloch’s theorem only applies to an infintely repeating potential and ourgoal is to model a macroscopic crystal so the number of repeating units ofthe potential is of the order of 1010
by assuming that the edges of the solid are always far away, and applyingperiodic boundary conditions, Bloch’s theorem can be made applicable tothe Dirac comb and other models of solids
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 8 / 14
Bloch functions
The eigenvalue equation for the displace-ment operator is thus
since λ cannot be zero, it must be a com-plex number which can expressed as aphasor
we thus have a condition on the wave-function in a periodic potential
resulting in Bloch’s theorem
Dψ(x) = λψ(x)
ψ(x + a) = λψ(x)
λ = e iKa
ψ(x + a)= e iKaψ(x)
Bloch’s theorem only applies to an infintely repeating potential and ourgoal is to model a macroscopic crystal so the number of repeating units ofthe potential is of the order of 1010
by assuming that the edges of the solid are always far away, and applyingperiodic boundary conditions, Bloch’s theorem can be made applicable tothe Dirac comb and other models of solids
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 8 / 14
Bloch functions
The eigenvalue equation for the displace-ment operator is thus
since λ cannot be zero, it must be a com-plex number which can expressed as aphasor
we thus have a condition on the wave-function in a periodic potential
resulting in Bloch’s theorem
Dψ(x) = λψ(x)
ψ(x + a) = λψ(x)
λ = e iKa
ψ(x + a)= e iKaψ(x)
Bloch’s theorem only applies to an infintely repeating potential and ourgoal is to model a macroscopic crystal so the number of repeating units ofthe potential is of the order of 1010
by assuming that the edges of the solid are always far away, and applyingperiodic boundary conditions, Bloch’s theorem can be made applicable tothe Dirac comb and other models of solids
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 8 / 14
Bloch functions
The eigenvalue equation for the displace-ment operator is thus
since λ cannot be zero, it must be a com-plex number which can expressed as aphasor
we thus have a condition on the wave-function in a periodic potential
resulting in Bloch’s theorem
Dψ(x) = λψ(x)
ψ(x + a) = λψ(x)
λ = e iKa
ψ(x + a)= e iKaψ(x)
Bloch’s theorem only applies to an infintely repeating potential and ourgoal is to model a macroscopic crystal so the number of repeating units ofthe potential is of the order of 1010
by assuming that the edges of the solid are always far away, and applyingperiodic boundary conditions, Bloch’s theorem can be made applicable tothe Dirac comb and other models of solids
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 8 / 14
Bloch functions
The eigenvalue equation for the displace-ment operator is thus
since λ cannot be zero, it must be a com-plex number which can expressed as aphasor
we thus have a condition on the wave-function in a periodic potential
resulting in Bloch’s theorem
Dψ(x) = λψ(x)
ψ(x + a) = λψ(x)
λ = e iKa
ψ(x + a)= e iKaψ(x)
Bloch’s theorem only applies to an infintely repeating potential and ourgoal is to model a macroscopic crystal so the number of repeating units ofthe potential is of the order of 1010
by assuming that the edges of the solid are always far away, and applyingperiodic boundary conditions, Bloch’s theorem can be made applicable tothe Dirac comb and other models of solids
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 8 / 14
Bloch functions
The eigenvalue equation for the displace-ment operator is thus
since λ cannot be zero, it must be a com-plex number which can expressed as aphasor
we thus have a condition on the wave-function in a periodic potential
resulting in Bloch’s theorem
Dψ(x) = λψ(x)
ψ(x + a) = λψ(x)
λ = e iKa
ψ(x + a)= e iKaψ(x)
Bloch’s theorem only applies to an infintely repeating potential and ourgoal is to model a macroscopic crystal so the number of repeating units ofthe potential is of the order of 1010
by assuming that the edges of the solid are always far away, and applyingperiodic boundary conditions, Bloch’s theorem can be made applicable tothe Dirac comb and other models of solids
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 8 / 14
Bloch functions
The eigenvalue equation for the displace-ment operator is thus
since λ cannot be zero, it must be a com-plex number which can expressed as aphasor
we thus have a condition on the wave-function in a periodic potential
resulting in Bloch’s theorem
Dψ(x) = λψ(x)
ψ(x + a) = λψ(x)
λ = e iKa
ψ(x + a)= e iKaψ(x)
Bloch’s theorem only applies to an infintely repeating potential and ourgoal is to model a macroscopic crystal so the number of repeating units ofthe potential is of the order of 1010
by assuming that the edges of the solid are always far away, and applyingperiodic boundary conditions, Bloch’s theorem can be made applicable tothe Dirac comb and other models of solids
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 8 / 14
Bloch functions
The eigenvalue equation for the displace-ment operator is thus
since λ cannot be zero, it must be a com-plex number which can expressed as aphasor
we thus have a condition on the wave-function in a periodic potential
resulting in Bloch’s theorem
Dψ(x) = λψ(x)
ψ(x + a) = λψ(x)
λ = e iKa
ψ(x + a)= e iKaψ(x)
Bloch’s theorem only applies to an infintely repeating potential and ourgoal is to model a macroscopic crystal so the number of repeating units ofthe potential is of the order of 1010
by assuming that the edges of the solid are always far away, and applyingperiodic boundary conditions, Bloch’s theorem can be made applicable tothe Dirac comb and other models of solids
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 8 / 14
Periodic boundary conditions
Take N to be the number of “atoms” in the system
wrap the end of the comb back to thestart to get a periodic boundary con-dition
this is just applying the displacementoperator N times and getting, byBloch’s theorem
leading to quantization of the factorK , which is clearly similar to the wavenumber in the free electron gas model
ψ(x + Na) = ψ(x)
e iNKaψ(x) = ψ(x)
e iNKa = 1
NKa = 2πn
K =2πn
Na, (n = 0,±1,±2, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 9 / 14
Periodic boundary conditions
Take N to be the number of “atoms” in the system
wrap the end of the comb back to thestart to get a periodic boundary con-dition
this is just applying the displacementoperator N times and getting, byBloch’s theorem
leading to quantization of the factorK , which is clearly similar to the wavenumber in the free electron gas model
ψ(x + Na) = ψ(x)
e iNKaψ(x) = ψ(x)
e iNKa = 1
NKa = 2πn
K =2πn
Na, (n = 0,±1,±2, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 9 / 14
Periodic boundary conditions
Take N to be the number of “atoms” in the system
wrap the end of the comb back to thestart to get a periodic boundary con-dition
this is just applying the displacementoperator N times and getting, byBloch’s theorem
leading to quantization of the factorK , which is clearly similar to the wavenumber in the free electron gas model
ψ(x + Na) = ψ(x)
e iNKaψ(x) = ψ(x)
e iNKa = 1
NKa = 2πn
K =2πn
Na, (n = 0,±1,±2, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 9 / 14
Periodic boundary conditions
Take N to be the number of “atoms” in the system
wrap the end of the comb back to thestart to get a periodic boundary con-dition
this is just applying the displacementoperator N times and getting, byBloch’s theorem
leading to quantization of the factorK , which is clearly similar to the wavenumber in the free electron gas model
ψ(x + Na) = ψ(x)
e iNKaψ(x) = ψ(x)
e iNKa = 1
NKa = 2πn
K =2πn
Na, (n = 0,±1,±2, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 9 / 14
Periodic boundary conditions
Take N to be the number of “atoms” in the system
wrap the end of the comb back to thestart to get a periodic boundary con-dition
this is just applying the displacementoperator N times and getting, byBloch’s theorem
leading to quantization of the factorK , which is clearly similar to the wavenumber in the free electron gas model
ψ(x + Na) = ψ(x)
e iNKaψ(x) = ψ(x)
e iNKa = 1
NKa = 2πn
K =2πn
Na, (n = 0,±1,±2, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 9 / 14
Periodic boundary conditions
Take N to be the number of “atoms” in the system
wrap the end of the comb back to thestart to get a periodic boundary con-dition
this is just applying the displacementoperator N times and getting, byBloch’s theorem
leading to quantization of the factorK , which is clearly similar to the wavenumber in the free electron gas model
ψ(x + Na) = ψ(x)
e iNKaψ(x) = ψ(x)
e iNKa = 1
NKa = 2πn
K =2πn
Na, (n = 0,±1,±2, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 9 / 14
Periodic boundary conditions
Take N to be the number of “atoms” in the system
wrap the end of the comb back to thestart to get a periodic boundary con-dition
this is just applying the displacementoperator N times and getting, byBloch’s theorem
leading to quantization of the factorK , which is clearly similar to the wavenumber in the free electron gas model
ψ(x + Na) = ψ(x)
e iNKaψ(x) = ψ(x)
e iNKa = 1
NKa = 2πn
K =2πn
Na, (n = 0,±1,±2, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 9 / 14
Periodic boundary conditions
Take N to be the number of “atoms” in the system
wrap the end of the comb back to thestart to get a periodic boundary con-dition
this is just applying the displacementoperator N times and getting, byBloch’s theorem
leading to quantization of the factorK , which is clearly similar to the wavenumber in the free electron gas model
ψ(x + Na) = ψ(x)
e iNKaψ(x) = ψ(x)
e iNKa = 1
NKa = 2πn
K =2πn
Na, (n = 0,±1,±2, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 9 / 14
Periodic boundary conditions
Take N to be the number of “atoms” in the system
wrap the end of the comb back to thestart to get a periodic boundary con-dition
this is just applying the displacementoperator N times and getting, byBloch’s theorem
leading to quantization of the factorK , which is clearly similar to the wavenumber in the free electron gas model
ψ(x + Na) = ψ(x)
e iNKaψ(x) = ψ(x)
e iNKa = 1
NKa = 2πn
K =2πn
Na, (n = 0,±1,±2, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 9 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions , where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we havedefining k =
√2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na)
= e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions , where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we havedefining k =
√2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na)
= e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions , where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we havedefining k =
√2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na) = e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions , where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we havedefining k =
√2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na) = e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions
, where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we havedefining k =
√2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na) = e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions
, where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we havedefining k =
√2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na) = e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions , where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we havedefining k =
√2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na) = e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions , where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we have
defining k =√
2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na) = e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions , where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we have
defining k =√
2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na) = e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions , where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we havedefining k =
√2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na) = e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb problem
Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit
the potential can be written as asum of delta functions , where αis the “strength” of the potentialsand N is the number of repeatingperiods
in the region 0 < x < a we havedefining k =
√2mE/~, gives
x-3a 0-2a-4a -a 2a 4a3aa
V(x)
. . .. . .
ψ(x) = ψ(x + Na) = e iNKaψ(x)
K =2πn
Na, (n = 0,±1,±2, . . . )
V (x) = α
N−1∑j=0
δ(x − ja)
Eψ = − ~2
2m
d2ψ
dx2
d2ψ
dx2= −k2ψ
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 10 / 14
Dirac comb solution
The general solution is one we have seen already
and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B
= e−iKa [A sin(ka) + B cos(ka)]
kA
− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already
and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B
= e−iKa [A sin(ka) + B cos(ka)]
kA
− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B
= e−iKa [A sin(ka) + B cos(ka)]
kA
− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B
= e−iKa [A sin(ka) + B cos(ka)]
kA
− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B
= e−iKa [A sin(ka) + B cos(ka)]
kA
− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B
= e−iKa [A sin(ka) + B cos(ka)]
kA
− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA
− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA
− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA
− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA− e−iKak [A cos(ka)− B sin(ka)]
= B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation
, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation
, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)
− e−iKak[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka)
+ e−iKak sin(ka) =2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka)
=2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
The general solution is one we have seen already and in the cellimmediately preceeding the origin, we can write the solution using Bloch’stheorem
ψ(x)= A sin(kx) + B cos(kx), (0 < x < a)
ψ(x)= e−iKa [A sin k(x + a) + B cos k(x + a)], (−a < x < 0)
continuity at x = 0 gives
while there is a discontinu-ity in the derivative
rearranging the continuityequation, substituting for A,and cancelling B
B = e−iKa [A sin(ka) + B cos(ka)]
kA− e−iKak [A cos(ka)− B sin(ka)] = B2mα
~2
A sin(ka) = B[e iKa − cos(ka)
]
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 11 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa
− cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka)
=2mα
~2ksin(ka)
e iKa
− 2 cos(ka) + e−iKa
=2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa
− cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka)
=2mα
~2ksin(ka)
e iKa
− 2 cos(ka) + e−iKa
=2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa
− cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka)
=2mα
~2ksin(ka)
e iKa
− 2 cos(ka) + e−iKa
=2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa
− cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka)
=2mα
~2ksin(ka)
e iKa
− 2 cos(ka) + e−iKa
=2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)
− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka)
=2mα
~2ksin(ka)
e iKa
− 2 cos(ka) + e−iKa
=2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka)
+ e−iKa cos2(ka) + e−iKa sin2(ka)
=2mα
~2ksin(ka)
e iKa
− 2 cos(ka) + e−iKa
=2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka)
+ e−iKa sin2(ka)
=2mα
~2ksin(ka)
e iKa
− 2 cos(ka) + e−iKa
=2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa
− 2 cos(ka) + e−iKa
=2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa
− 2 cos(ka) + e−iKa
=2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − 2 cos(ka)
+ e−iKa
=2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − 2 cos(ka) + e−iKa =2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − 2 cos(ka) + e−iKa =2mα
~2ksin(ka)
2 cos(Ka)
− 2 cos(ka)
=2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − 2 cos(ka) + e−iKa =2mα
~2ksin(ka)
2 cos(Ka)− 2 cos(ka) =2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − 2 cos(ka) + e−iKa =2mα
~2ksin(ka)
2 cos(Ka)− 2 cos(ka) =2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − 2 cos(ka) + e−iKa =2mα
~2ksin(ka)
2 cos(Ka)− 2 cos(ka) =2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify
cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − 2 cos(ka) + e−iKa =2mα
~2ksin(ka)
2 cos(Ka)− 2 cos(ka) =2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) =
cos(z) + βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − 2 cos(ka) + e−iKa =2mα
~2ksin(ka)
2 cos(Ka)− 2 cos(ka) =2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) = cos(z)
+ βsin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − 2 cos(ka) + e−iKa =2mα
~2ksin(ka)
2 cos(Ka)− 2 cos(ka) =2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) = cos(z) + β
sin(z)
z
= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
k[e iKa − cos(ka)]
sin(ka)− e−iKak
[e iKa − cos(ka)]
sin(ka)cos(ka) + e−iKak sin(ka) =
2mα
~2
[e iKa − cos(ka)][1− e−iKa cos(ka)] + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − cos(ka)− cos(ka) + e−iKa cos2(ka) + e−iKa sin2(ka) =2mα
~2ksin(ka)
e iKa − 2 cos(ka) + e−iKa =2mα
~2ksin(ka)
2 cos(Ka)− 2 cos(ka) =2mα
~2ksin(ka)
cos(Ka) = cos(ka) +mα
~2ksin(ka)
letting z ≡ ka and β = mαa/~2 wesimplify cos(Ka) = cos(z) + β
sin(z)
z= f (z)
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 12 / 14
Dirac comb solution
cos(Ka) = cos(z) + βsin(z)
z
-1
0
+1
0 π 2π 3π 4π
f(z)
z
Since | cos(Ka)|≤1, the shaded re-gions contain the only valid solu-tions and are called “bands”
the “gaps” are forbidden energies,whose values of k are not allowed
since K = 2πn/Na is quantized(n = 0,±1,±2, . . . ), so are k andE in the allowed intervals
there are N allowed states in eachband but the the energies are nearlycontinuous since N is very large
the bands and gaps are able to ex-plain the properties of metals, semi-conductors and insulators in a sim-ple way
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 13 / 14
Dirac comb solution
cos(Ka) = cos(z) + βsin(z)
z
-1
0
+1
0 π 2π 3π 4π
f(z)
z
Since | cos(Ka)|≤1, the shaded re-gions contain the only valid solu-tions and are called “bands”
the “gaps” are forbidden energies,whose values of k are not allowed
since K = 2πn/Na is quantized(n = 0,±1,±2, . . . ), so are k andE in the allowed intervals
there are N allowed states in eachband but the the energies are nearlycontinuous since N is very large
the bands and gaps are able to ex-plain the properties of metals, semi-conductors and insulators in a sim-ple way
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 13 / 14
Dirac comb solution
cos(Ka) = cos(z) + βsin(z)
z
-1
0
+1
0 π 2π 3π 4π
f(z)
z
Since | cos(Ka)|≤1, the shaded re-gions contain the only valid solu-tions and are called “bands”
the “gaps” are forbidden energies,whose values of k are not allowed
since K = 2πn/Na is quantized(n = 0,±1,±2, . . . ), so are k andE in the allowed intervals
there are N allowed states in eachband but the the energies are nearlycontinuous since N is very large
the bands and gaps are able to ex-plain the properties of metals, semi-conductors and insulators in a sim-ple way
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 13 / 14
Dirac comb solution
cos(Ka) = cos(z) + βsin(z)
z
-1
0
+1
0 π 2π 3π 4π
f(z)
z
Since | cos(Ka)|≤1, the shaded re-gions contain the only valid solu-tions and are called “bands”
the “gaps” are forbidden energies,whose values of k are not allowed
since K = 2πn/Na is quantized(n = 0,±1,±2, . . . ), so are k andE in the allowed intervals
there are N allowed states in eachband but the the energies are nearlycontinuous since N is very large
the bands and gaps are able to ex-plain the properties of metals, semi-conductors and insulators in a sim-ple way
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 13 / 14
Dirac comb solution
cos(Ka) = cos(z) + βsin(z)
z
-1
0
+1
0 π 2π 3π 4π
f(z)
z
Since | cos(Ka)|≤1, the shaded re-gions contain the only valid solu-tions and are called “bands”
the “gaps” are forbidden energies,whose values of k are not allowed
since K = 2πn/Na is quantized(n = 0,±1,±2, . . . ), so are k andE in the allowed intervals
there are N allowed states in eachband but the the energies are nearlycontinuous since N is very large
the bands and gaps are able to ex-plain the properties of metals, semi-conductors and insulators in a sim-ple way
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 13 / 14
Dirac comb solution
cos(Ka) = cos(z) + βsin(z)
z
-1
0
+1
0 π 2π 3π 4π
f(z)
z
Since | cos(Ka)|≤1, the shaded re-gions contain the only valid solu-tions and are called “bands”
the “gaps” are forbidden energies,whose values of k are not allowed
since K = 2πn/Na is quantized(n = 0,±1,±2, . . . ), so are k andE in the allowed intervals
there are N allowed states in eachband but the the energies are nearlycontinuous since N is very large
the bands and gaps are able to ex-plain the properties of metals, semi-conductors and insulators in a sim-ple way
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 13 / 14
Dirac comb solution
cos(Ka) = cos(z) + βsin(z)
z
-1
0
+1
0 π 2π 3π 4π
f(z)
z
Since | cos(Ka)|≤1, the shaded re-gions contain the only valid solu-tions and are called “bands”
the “gaps” are forbidden energies,whose values of k are not allowed
since K = 2πn/Na is quantized(n = 0,±1,±2, . . . ), so are k andE in the allowed intervals
there are N allowed states in eachband but the the energies are nearlycontinuous since N is very large
the bands and gaps are able to ex-plain the properties of metals, semi-conductors and insulators in a sim-ple way
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 13 / 14
Types of solidsE
k
Plotting these “bands” on an energy plotshows that there are 2N closely spacedstates in each of the shaded regions sepa-rated by “gaps” where there are no avail-able states.
The low-lying states are localized in en-ergy, higher states are more spread out.
If a material has an even number of elec-trons per atom, then the topmost bandwith electrons is completely filled, givingan insulator.
If a material is made up of atoms whichhave an odd number of electrons, the top-most band with electrons is half filled, giv-ing a metal.
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 14 / 14
Types of solidsE
k
Plotting these “bands” on an energy plotshows that there are 2N closely spacedstates in each of the shaded regions sepa-rated by “gaps” where there are no avail-able states.
The low-lying states are localized in en-ergy, higher states are more spread out.
If a material has an even number of elec-trons per atom, then the topmost bandwith electrons is completely filled, givingan insulator.
If a material is made up of atoms whichhave an odd number of electrons, the top-most band with electrons is half filled, giv-ing a metal.
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 14 / 14
Types of solidsE
k
Plotting these “bands” on an energy plotshows that there are 2N closely spacedstates in each of the shaded regions sepa-rated by “gaps” where there are no avail-able states.
The low-lying states are localized in en-ergy, higher states are more spread out.
If a material has an even number of elec-trons per atom, then the topmost bandwith electrons is completely filled, givingan insulator.
If a material is made up of atoms whichhave an odd number of electrons, the top-most band with electrons is half filled, giv-ing a metal.
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 14 / 14
Types of solidsE
k
Plotting these “bands” on an energy plotshows that there are 2N closely spacedstates in each of the shaded regions sepa-rated by “gaps” where there are no avail-able states.
The low-lying states are localized in en-ergy, higher states are more spread out.
If a material has an even number of elec-trons per atom, then the topmost bandwith electrons is completely filled, givingan insulator.
If a material is made up of atoms whichhave an odd number of electrons, the top-most band with electrons is half filled, giv-ing a metal.
C. Segre (IIT) PHYS 405 - Fall 2015 November 18, 2015 14 / 14
top related